Use the mirror equation to deduce that :
d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note : The exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

Explanation

The mirror equation is `(1)/(u)+(1)/(v)=(1)/(f)`
But `m=(upsilon)/(u)=(f)/(u-f)=(upsilon-f)/(f)`
Hence `m=(f)/(u-f)`
`upsilon =f(m+1)`
d) When `u gt 0 lt f` we get
`m=(f)/(u-f)=(f)/((gt 0 lt f)-f)=(f)/((gt -f lt 0))`
`= gt -1`
(`because` m is negative, image is virtual and enlarged because is numerically `y gt 1`).